$f(U_1|U_1>U_2)$ where $U_1$ and $U_2$ are independent uniform variables We have two distributions of uniform random variables over $(0, 1)$ named $U_1$ and $U_2$ (and they are independent). How can we calculate $f(U_1|U_1>U_2)$, $f(U_2|U_1>U_2)$, $E(U_1|U_1>U_2)$ and $E(U_2|U_1>U_2)$?
I tried to use $f(x|y) = f(x,y)/f(y)$ but what does $f(U_1, U_1>U_2)$ even means? I can't think about it since $U_1>U_2$ is a condition and $U_1$ is a random variable.
 A: On a general basis [rather in the special case you mention, since this seems to come from an exercise], finding the distribution of one or several random variable(s) $X$ given an event $A$ can proceed by introducing the event as a new random variable$$Y=\mathbb{I}_A(X)$$which takes values 1 and 0 depending on whether or not the event occurred. This means creating the joint distribution of $X$ and $Y$, which can be derived from the distribution density $f$ of $X$ [assuming a density with respect to a given measure] as$$X,Y\sim f(x)\times \mathbb{I}_A(x)^y \mathbb{I}_{A^c}(x)^{1-y}$$From the joint density, you can derive the conditional$$X|Y=y\sim f(x)\times \mathbb{I}_A(x)^y \mathbb{I}_{A^c}(x)^{1-y}\big/ \mathbb{P}(A)^y\mathbb{P}(A^c)^{1-y}$$
A: The picture below may help you to get some intuition behind the meaning of $f(U_1\vert U_1>U_2)$ and $f(U_1 , U_1>U_2)$ in terms of relative area.


You could evaluate the probability by using an integral that computes the cumulative distribution function. 
$$P(U_1 \leq x \vert U_1>U_2) = \frac{P(U_1 \leq x,  U_1>U_2)}{P(U_1>U_2)} = \frac{\int_0^x \left( \int_{U_2}^x f(U_1,U_2) dU_1 \right) dU_2}{\int_0^1 \left( \int_{U_2}^1 f(U_1,U_2) dU_1 \right) dU_2 } = x^2$$
where the second equality follows when we plug in the uniform density $f(U_1,U_2)=1$. 

To go back to the intuitive picture. This $P(U_1 \leq x , U_1>U_2)$ is related to the area $\frac{1}{2}x^2$ of the little triangle from (0,0)-(0,x)-(x,x), which is when $U_1<x$ and $U_1>U2$. This $P(U_1>U_2)$ is related to the area $\frac{1}{2}$ of the triangle from (0,0)-(0,1)-(1,1), which is when $U_1<1$ and $U_1>U_2$.
